can "using namespace std;" and "std::cout" be used together? - c++

No an error occurs when using using namespace std; and std::cout together. Can these two be used together?
#include <iostream>
using namespace std;
int main() {
std::cout << "Hello world!";
return 0;
}

There is no problem. In this statement
std::cout << "Hello world!";
there is used the qualified name lookup of the name cout in the namespace std.
You could also write
cout << "Hello world!";
and in this case there would be used the unqualified name lookup and the name cout would be found due to the directive
using namespace std;
You could also include the using declaration like
using std::cout;
Pay attention to that you should avoid to use the using directive. It can be a reason of ambiguity. It is much better to use qualified names.

The purpose of namespace is to protect against collision. When you type in using namespace std; it turns off that protection. This lets you use cout, string, vector... without std:: resolution, but they may collide with other namespaces.
In some tutorials you may see using namespace std; They put that in there to make the examples shorter without having to type in std:: every where. But that usage is limited to short examples. For actual code it is recommended not to add using namespace std;
You can always use std::cout, std::string, std::vector etc. without worrying about collision.

Related

Why do I need to do 'using namespace std' instead of 'using std::cout'?

everyone.
I have learnt that it is often desirable to write in my codes using std::cout instead of using namespace std in order to avoid namespace conflicts. In the following script I only use cout and if I write std:: cout instead of using namespace std; it does not work.
Can anyone please help me understand why? In general, when does std::cout not work and I am forced to use using namespace std?
#include <iostream>
#include <string>
using std::cout; //if writing here "using namespace std;" instead, the code does not work
class MyClass{
public:
string name;
MyClass (string n)
{
name=n;
cout<<"Hello "<<name;
}
};
int main()
{
MyClass MyObj("Mike");
return 0;
}
You need to add using std::string; along with using std::cout; to make it work as you're not only using cout from namespace std, string is also a member of namespace std which you are using in your code.
Your code works okay with:
using std::cout;
statement for cout, but the compiler must know the location of string (actually it's std::string) too which you're currently using. You must define:
using std::string;
When you enter:
using namespace std;
It calls the entire namespace called std which contains a variety of features added as C++ standard library and then you don't need to use prefix std:: for those functions/classes/variables of that namespace.

What does using namespace std; do?

What is the matter with this C++ code?
I get an error message saying: that there should be a ; before z, which is not correct. The only part that I don't understand is the purpose line 2 serves.
#include <iostream>
using namespace std;
int subtraction(int a, int b)
{
int r;
r=a-b;
return r;
}
int main()
{
int z;
z = subtraction (5,9);
cout z;
}
Thank you in advance.
using namespace std; means you can write cout later on rather than std::cout. It saves typing at the expense of gross namespace pollution.
Your compile error can be fixed by writing cout << z;
Also, do return a value from main.
To begin to explain what this does, it is important for you to understand what namespaces do. Namespaces are logical units of code divided up, you are able to create your own namespace or use other namespaces. The benefit of using namespaces is to have your program logically divide up your code with like functions. This is very similar to classes, but you do not need initiation of the namespaces like classes do. IN Java this would be similar to packages. To use a function within a namespace you need to use the namespace identifier followed by the function you are calling. This will call the correct function in the namespace scope you are wanting to use. An example of creating a namespace is the following:
namespace connection
{
int create_connection();
int close_connection();
//ect.......
}
Then later in the code when you want to call create_connection you need to do it the following way:
connection::create_connection();
As for your answer you are able to prevent from having to type the namespace identifier in this case connection, or in your case std. You are able to introduce an entire namespace into a section of code by using a using-directive. This will allow you to call the functions that are in that namespace without needing to use the namespace followed with the scope indicator" :: ".
The following syntax to do this is as follows:
using namespace connection:
or in your case
using namespace std;
So by doing this with std you will be granting that access to std namespace which includes C++ I/O objects cout and cin to use freely without having to use the namespace and scope operator first. Though a better practice is to limit the scope to the namespace members you want to actually use. On large programs this will be cleaner why of coding as well as avoid several problems. To introduce only specific members of a namespace, such as only introducing the std::cin and std::cout, you do the following:
using std::cin;
using std::cout;
What does using namespace std; do?
It tells the compiler which class/namespace to look in for an identifier. You either use using namespace std; in the beginning of the file or have to place it in front of each function that belongs to it.
What is the matter with this C++ code?
The syntax to use std::cout is:
std::cout << source;
variable source is inserted with the help of operator << in the std::cout stream which prints it to the standard output, i.e. computer monitor.
std "labels" a function member of the Standard Library. This is a technique used (among other reasons) to resolve (using the resolution operator ::) members that belong to the Standard Library from (possible) name conflicts with functions with similar(same) names and to reduce a scope of a search. std is called a namespace, so using namespace std; is a bit self explanatory.

About std:cout in C++

Is there an error in this code:
#include <iostream>
using namespace std;
int main()
{
std:cout << "hello" "\n";
}
GCC detects no error but std:cout does not seem standard.
There's no error. I could rewrite your code to make it clearer:
#include <iostream>
using namespace std;
int main()
{
std:
cout << "hello" "\n";
}
You created a label named std. cout is used unqualified, which is okay since you have the using-directive for std above it. And you can concatenate string literals by writing them next to each other as you did. This is perfectly well-formed code that prints "hello" followed by a newline.
You're defining a label std and then you're calling cout. This is legal because you have using namespace std;
The code has an issue.
while trying to instruct the compiler to use the namespace std, we are trying to call the function cout which is defined in the scope of std.
thus the correct use of the scope resolution operator is
'std::cout '
and not
std:cout.
And others have pointed out
by writing
std:
what you do is create a label.

Using Namespace std

I am taking a programming class in school and I wanted to start doing some c++ programming out of class. My school using Microsoft Visual C++ 6.0 (which is from 1998) so it still uses <iostream.h> rather than <iostream> and using namespace std. When I started working, I couldn't figure out how and when to use using namespace std and when to just use things like std::cout<<"Hello World!"<<'\n'; (for example) as well as it's limits and other uses for the namespace keyword. In particular, if I want to make a program with iostream and iomanip, do I have to state "using namespace std" twice, or is there something different that I would have to use as well, or can I just do the same thing as I did with iostream? I tried googling it but I didn't really understand anything. Thanks in advance for the help.
Ok, handful of things there, but it is manageable.
First off, the difference between:
using namespace std;
...
cout << "Something" << endl;
And using
std::cout << "Something" << std::endl;
Is simply a matter of scope. Scope is just a fancy way of saying how the compiler recognizes names of variables and functions, among other things. A namespace does nothing more than add an extra layer of scope onto all variables within that namespace. When you type using namespace std, you are taking everything inside of the namespace std and moving it to the global scope, so that you can use the shorter cout instead of the more fully-qualified std::cout.
One thing to understand about namespaces is that they stretch across files. Both <iostream> and <iomanip> use the namespace std. Therefore, if you include both, then the declaration of using namespace std will operate on both files, and all symbols in both files will be moved to the global scope of your program (or a function's scope, if you used it inside a function).
There are going to be people who tell you "don't use using namespace std!!!!", but they rarely tell you why. Lets say that I have the following program, where all I am trying to do is define two integers and print them out:
#include <iostream>
using namespace std;
int main(int argc, char** argv) {
int cout = 0;
int endl = 1;
cout << cout << endl << endl; // The compiler WILL freak out at this :)
return 0;
}
When I use using namespace std, I am opening the door for naming collisions. If I (by random chance), have named a variable to be the same thing as what was defined in a header, then your program will break, and you will have a tough time figuring out why.
I can write the same program as before (but get it to work) by not using the statement using namespace std:
#include <iostream>
int main(int argc, char** argv) {
int cout = 0;
int endl = 1;
std::cout << cout << endl << std::endl; // Compiler is happy, so I'm happy :)
return 0;
}
Hopefully this has clarified a few things.
If you use the header names without the .h, then the stuff declared/defined in it will be in the std namespace. You only have to use using namespace std; once in the scope where you want stuff imported in order to get everything; more than one using namespace std; doesn't help anything.
I'd recommend against using namespace std; in general, though. I prefer to say, for example, using std::cout; instead, in order to keep names in std from conflicting with mine.
For example:
#include <iostream>
#include <iomanip>
int main()
{
using namespace std;
int left = 1, right = 2;
cout << left << " to " << right << "\n";
}
may cause mysterious issues, because left and right exist in the std namespace (as IO manipulators), and they get imported if you lazily say using namespace std;. If you meant to actually use the IO manipulators rather than output the variables, you may be a bit disappointed. But the intent isn't obvious either way. Maybe you just forgot you have ints named left and right.
Instead, if you say
#include <iostream>
#include <iomanip>
int main()
{
using std::cout;
int left = 1, right = 2;
cout << left << " to " << right << "\n";
}
or
#include <iostream>
#include <iomanip>
int main()
{
int left = 1, right = 2;
std::cout << left << " to " << right << "\n";
}
everything works as expected. Plus, you get to see what you're actually using (which, in this case, includes nothing from <iomanip>), so it's easier to keep your includes trimmed down to just what you need.
Here is a good link that describes namespaces and how they work.
Both methods are correct, that is, you can either introduce a namespace with the "using" statement or you can qualify all the members of the namespace. Its a matter of coding style. I prefer qualifying with namespaces because it makes it clear to the reader in which namespace the function / class is defined.
Also, you do not have to introduce a namespace twice if you are including multiple files. One using statement is enough.
Good question, Ryan. What using namespace does is importing all symbols from a given namespace (scope) into the scope where it was used. For example, you can do the following:
namespace A {
struct foo {};
}
namespace B {
using namespace A;
struct bar : foo {};
}
In the above examples, all symbols in namespace A become visible in namespace B, like they were declared there.
This import has affect only for a given translation unit. So, for example, when in your implementation file (i.e. .cpp) you do using namespace std;, you basically import all symbols from std namespace into a global scope.
You can also import certain symbols rather than everything, for example:
using std::cout;
using std::endl;
You can do that in global scope, namespace scope or function scope, like this:
int main ()
{
using namespace std;
}
It is up to a programmer to decide when to use fully qualified names and when to use using keyword. Usually, it is a very bad taste to put using into a header files. Professional C++ programmers almost never do that, unless that is necessary to work around some issue or they are 100% sure it will not mess up type resolution for whoever use that header.
Inside the source file, however (nobody includes source files), it is OK to do any sort of using statements as long as there are no conflicting names in different namespaces. It is only a matter of taste. For example, if there are tons of symbols from different namespaces being used all over the code, I'd prefer at least some hints as for where they are actully declared. But everyone is familiar with STL, so using namespace std; should never do any harm.
There also could be some long namespaces, and namespace aliasing comes handy in those cases. For example, there is a Boost.Filesystem library that puts all of its symbols in boost::filesystem namespace. Using that namespace would be too much, so people usually do something like this:
namespace fs = boost::filesystem;
fs::foo ();
fs::bar ();
Also, it is almost OK to use namespace aliasing in headers, like this:
namespace MyLib {
namespace fs = boost::filesystem;
}
.. and benefit from less typing. What happens is that users that will use this header, will not import the whole filesystem library by saying using namespace MyLib;. But then, they will import "fs" namespace from your library that could conflict with something else. So it is better not to do it, but if you want it too badly, it is better than saying using namespace boost::filesystem there.
So getting back to your question. If you write a library using C++ I/O streams, it is better not to have any using statements in headers, and I'd go with using namespace std; in every cpp file. For example:
somefile.hpp:
namespace mylib {
class myfile : public std::fstream {
public:
myfile (const char *path);
// ...
};
}
somefile.cpp:
#include "somefile.hpp"
using namespace std;
using namespace mylib;
myfile::myfile (const char *path) : fstream (path)
{
// ...
}
Specific to using namespace std
You really shouldn't ever use it in a header file. By doing so, you've imported the entire 'std' into the global namespace for anyone who includes your header file, or for anyone else that includes a file that includes your file.
Using it inside a .cpp file, that's personal preference. I typically do not import the entire std into the global namespace, but there doesn't appear to be any harm in doing it yourself, to save a bit of typing.

Namespace using declaration (C++ Primer - Stanley Lipmann)

Anyone can help me to understand this statement found in chapter 3 (Library Types) by Stanley Lipmann?
"Using an unqualified version of a namespace name without a using declaration is an error, although some compilers may fail to detect this error"
I'm having such hard time understanding the semantics of his sentence (english).
Is he trying to say something like the below scenario?
int main() {
xx::yy
}
where xx is a namespace not defined using the "using" statement and yy is a member?
Example:
cout is a name of the std namespace. The unqualified name is cout. The qualified name is std::cout. It is an error to use the unqualified name(cout) without a using declaration beforehand. You can use either one of the two following declarations:
// This brings in the entire std namespace
using namespace std;
OR
// This only brings in cout. You would still need to qualify other names,
// such as cin, endl, etc...
using std::cout;
What he's saying is that the following code should not compile:
#include <iostream>
void foo() {
cout << "This is an error!" << endl;
}
The cout and endl names are not defined right now. They're declared as std::cout and std::endl, and in order to use them, you can do one of a few things:
#include <iostream>
void foo() {
std::cout << "This, I think, is the best way to do it." << std::endl;
}
Using the fully qualified name prevents collisions later on: you'll never have something else called std::cout.
#include <iostream>
void foo() {
using std::cout;
using std::endl;
cout << "This is pretty good." << endl;
}
Having the using statements specify the exact names you're using, and having the using statements in the function, can save some typing and makes collisions pretty unlikely.
#include <iostream>
using namespace std;
void foo() {
cout << "This works, but isn't good." << endl;
}
Importing the entire std namespace makes it pretty likely that you'll end up having a function named the same as an std function.. You might discover that as soon as you write it, or you might write your function and then later include the header file with the std version of the function, at which point your application will mysteriously break.
A namespace name is the name of a namespace.
namespace A {
}
namespace B = A;
The statement says that using a namespace name without a using declaration is an error. But that's not true: The above code is fine, still using the namespace-name A as an unqualified name.
Probably it should say the following, to convey its meaning
"Using an unqualified version of a namespace member name without a using declaration
outside the scope of the namespace is an error, although some compilers may fail to
detect this error"
Mentioning the scope is important. The following, for example, is fine too, even though it uses the unqualified version of a namespace member name
namespace A {
int x;
int &y = x; // x is an unqualified name
}
Books should be careful to try and not use slippery language. And even outside the scope of the namespace, the above sentence is not entirely correct because you can also extend the scope of x by a using directive. Using declarations aren't the only way to name a namespace member outside the namespace using an unqualified name.